Find the six trigonometric functions that correspond to the point on the unit circle

Question

FIND the six trigonometric functions that correspond to the unit circle

(4/5,3/5)

 

I dont even know were to start

Answer 1

For any point, P(x,y) := P(r cos(θ), r sin(θ)), where r = √((x)^2+(y)^2), and θ is an angle with the positive x-axis its initial ray and with a terminal ray starting at the origin and passing through P. [ The symbol ":=" means "is defined as".]

 

 

So, for A(4/5,3/5) := A(r cos(θ), r sin(θ)), where r = √((4/5)^2+(3/5)^2) = 1.

 

We pull sine and cosine out of A and use them to find the others.

 

sin(θ) = 3/5

cos(θ) = 4/5

tan(θ) := sin(θ)/cos(θ) = 3/4

cot(θ) := 1/tan(θ) = 4/3

sec(θ) := 1/cos(θ) = 5/4

csc(θ) := 1/sin(θ) = 5/3

Answer 2

Here you have aright triangle with 3, 4,5 

  5 is magnitude of the Hypotenuse , 3 , 4 magnitudes of the legs.

 

     Placing a triangle similar to this in unit circle.

 

        OAH , O is the center of the circle A , point on the circle, and B is the perpendicular from A to radius

 

   Cos A°=  Sin O° =  AH/ OA = 3/5 = 0.6/1

    Sin A = Cos O° = OH / OA = 4/5 = 0.8/1

    Cot A°= Tan O° = AH/ OH = 3/ 4 = 0.75/1

    Sec A° = CSC O° = OA/ AH = 5/3 = 1/0.6

    CSC A° = SEC O° = OA/ OH= 5/4 = 1/ 0.8

 

      

Answer 3

This is a good time to recall the Pythagorean Theorem. x^2 + y^ 2 = R^2

where (x/R , y/R ) = (4/5 , 3/5)

 

we can verify this is on the unit circle because  x^2 + y^ 2 = R^2 is true.

cos(theta) = 4/5

sin(theta) = 3/5

tan(theta) = 3/4

sec(theta) = 5/4

csc(theta) = 5/3

cot(theta) = 4/3

 

The answer will depend whether they want theta in rad or degrees, but those expressions above correspond to the circle.